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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Exam 1 -- October 10, 2003}
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{\it Directions}   Do all work in the blue exam booklet.  There are 
100 regular points and 10 extra credit points.
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I.  (20) A manufacturer of electronic components tests the lifetimes
of a certain type of battery and finds the following data:
$$123, 116, 122, 110, 125, 126, 111, 118, 117, 120$$
(lifetimes in hours).
How many of the sample points are within one standard deviation of 
the sample mean?  Is there reason to believe the lifetime
of this type of battery is not normally distributed from this small sample?
Explain.
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{\it Solution:}  The sample mean is 
$$\overline{x} = {1\over 10}\sum_{i=1}^{10} x_i = 118.8$$
The sample standard deviation is
$$s = \sqrt{{1\over 9}\sum_{i=1}^{10} (x_i - \overline x)^2} \doteq 5.47$$
There are exactly 6 out of the 10 data points within 1 standard deviation of 
the mean:  $123,116,122,118,117,120$.  This is quite close to 
the $68\%$ we would expect for a normal distribution.  Especially
with such a small sample, there is no reason to suspect that
the battery lifetimes are not normally distributed.
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II.  (15) In a regional spelling bee, the 10 finalists consist
of 5 girls and 5 boys.  Assume that all the finalists are equally 
proficient spellers and that the outcome of the contest is random.
What is the probability that 4 of the top 5 finishers will be female?
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{\it Solution:}  This can be done either by considering the 
five top finishers as an ordered list or as an unordered list -- 
there are exactly $5!$ orderings and that ``cancels out'' in 
the ratio for the probability.  If we think of the five top finishers 
as a ``committee'' (unordered) of 4 girls and 1 boy selected from the 
10 finalists, the probability is ${5\choose 4}{5\choose 1}/{10 \choose 5}$.
(Note this is one value of a hypergeometric pdf!). 
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III.  
\item{A)}  (15) State and prove the Law of Total Probability.
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{\it Solution}: See class notes.
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\item{B)}  (10) The Podunk City police department plans a crackdown
on speeders by placing radar traps at four different locations
$L_1,L_2,L_3,L_4$.  The probability that each of the 
traps is manned at any one time is $.4, .3, .2, .3$ respectively
(and the police really mean business -- if a trap is manned every speeder 
who passes it will get a ticket).  Speeders have probabilities 
of passing the four locations of $.2, .1, .5, .2$ respectively, 
and no one passes more than one.  What is the 
probability that a given speeder will actually receive a ticket?
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{\it Solution:}  Let $T$ be the event that the speeder gets a 
ticket, and let $L_i$ be the event that the speeder passes location $i$.
The second set of given information $.2,.1,.5.,2$ are the $P(L_i)$.
The first are $P(T|L_i)$ since the speeder gets a ticket whenever
he passes that trap and the trap is manned.  Every speeder passes one
and only one of the four trap locations, so they form the partition of 
the sample space of speeders.  By the Law of Total Probability, we
have
$$P(T) = \sum_{i=1}^4 P(T|L_i)P(L_i) = (.4)(.2)+(.3)(.1)+(.2)(.5)+(.3)(.2) = .27$$ 
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\item{C)}  (10) In the situation of part B, given 
that a speeder received a ticket, what
is the probability that he passed location $L_2$?
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{\it Solution:}  By Bayes' Rule,
$$P(L_2|T) = P(T|L_2)P(L_2)/.27 = (.3)(.1)/(.27) = .\overline{1} = 1/9.$$
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IV.  (10) Let $A,B$ be events for which $P(A) = .2$, $P(B) = .3$ and
$P(A\cap B) = .06$.  Are $\overline{A}$ and $\overline{B}$ independent
events?
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{\it Solution:} By the complement rule, $P(\overline{A})=.8$ and 
$P(\overline{B}) = .7$.  By the sum rule 
$P(A\cup B) = P(A) + P(B) - P(A\cap B) = .2 + .3 - .06 = .44$.
Hence $P(\overline{A\cup B})=P(\overline{A} \cap \overline{B}) =  1-.44 = .56$.
Since $.56 = (.8)(.7)$, $\overline{A}$ and $\overline{B}$ {\it are 
independent events}.
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V.  An allergist knows that $30\%$ of all people are 
allergic to the pollen of burdock weed.
\item{A)}  (10)  Starting from some point in time, the first
five patients in sequence that 
the allergist sees are not allergic to burdock pollen.  
Given that, what is the probability that the first patient
the doctor sees who is allergic to burdock pollen will come
after the 20th patient seen?  Explain the assumptions you are making 
to derive your solution.
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{\it Solution:}  Since the problem asks about the {\it number in sequence
of the first allergic patient}, this situation is dealing with
a {\it geometric} random variable $Y$.  The question translates
into: What is the conditional probability
$P(Y > 20 | Y > 5)$?  For the geometric random variable
with $p = .3$, $p(y) = (.7)^{y-1}(.3)$, so using the formula
for the sum of a geometric series,
$$\eqalign{
P(Y>20 | Y>5) &= {\sum_{y=21}^\infty (.7)^{y-1}(.3)\over\sum_{y=6}^\infty (.7)^{y-1}(.3)}\cr
              &= {(.7)^{20} \over (.7)^5}\cr
              &= .7^{15}\cr}$$
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\item{B)}  (10) What is the probability that from
4 to 7 (inclusive) of the next 20 patients she sees are allergic to 
burdock pollen?  Explain the assumptions you are making to derive 
your solution.
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{\it Solution:}  Since the question is asking about
the number of ``successes'' out of 20 independent ``trials'' (the
patients), here we are dealing with a binomial random variable $Y$
with $n = 20$, $p = .3$.  Using the table of binomial probabilities,
$$P(4 \le Y \le 7) = P(Y \le 7) - P(Y \le 3) = .772 - .107 = .665$$
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{\it Extra Credit} (10)  The Yankees won 10 of the 19 regular season games 
they played against the Red Sox this year.  Assuming that the Yankees
have a $10/19$ chance of winning each game they play against
the Sox, as of last Tuesday (i.e. before the first game), what 
was the probability that the Sox will clinch the ALCS at home?
Assume the games are independent events,  and recall that
the series starts with two games in New York, then moves to 
Boston for the next 3 (if needed),
then back to New York for the remainder of the series (if needed).
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{\it Answer:}  $\left({9\over 19}\right)^4 + 4 \left({9\over 19}\right)^4 {10\over 19}$
\bye
